Based on the shape of the curve shown in this post and the possible relationship shown here, I am willing to venture a conjecture:
$$\frac{\sigma (n)}{e^{\gamma} n \log \log n}<1-\frac{0.242692}{\ln(n)}$$
This conjecture holds for the largest Colossally Abundant Number I could find, the $143215^{th}$ with over 800k digits computed by Schwabhäuser. From his statistics:
$$ \frac{X(n_{143215})}{e^\gamma}=0.99995934<1-\frac{0.242692}{\ln(n_{143215})}=0.999999873$$
Can this bound or something similar be proven to hold for all Superabundant Numbers (and therefore all numbers)?